I was thinking of the example in Folland on page 61 i.e. &#x03BC;<!-- μ --> ( <mrow class=

Mackenzie Rios

Mackenzie Rios

Answered question

2022-05-23

I was thinking of the example in Folland on page 61 i.e. μ ( R ) = . Let f n = n χ [ 0 , 1 / n ] 0 a.e. Then f n 0 in measure. My inclination is that this is true so it requires proof. Also the same question is asked but for when μ ( X ) = 1. My inclination is that this is not true and a counterexample can be provided.

Answer & Explanation

Annabella Velez

Annabella Velez

Beginner2022-05-24Added 7 answers

The statement that convergence a.e. implies convergence in measure is true in finite measure spaces, not true in general.
For the first statement, note f n f a.e. implies for all ε > 0 and x X, there exists n such that k n implies | f n ( x ) f ( x ) | < ε. In particular, this implies
n k n { x X : | f k ( x ) f ( x ) | ε } = .
Finiteness of the measure space gives lets us use continuity from above to yield
lim n μ ( k n { x X : | f k ( x ) f ( x ) | ε } ) = 0
which implies convergence in measure.
To see how this fails in general measure spaces, consider f n = 1 ( , n ] , which converges to 1 everywhere but not in measure.

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